BZOJ 3295 动态逆序对

Posted 2020-08-09 ziliuziliu

调了好久。。。。

转化成三维偏序，cdq处理。

好像比较快？

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 300500
using namespace std;
long long n,m,x,a[maxn],pos[maxn],f1[maxn],f2[maxn],t[maxn],ans=0;
struct pnt
{
    long long a,b,c,ans;
}p[maxn];
bool cmp1(pnt x,pnt y) {return x.b<y.b;}
bool cmp2(pnt x,pnt y) {return x.b>y.b;}
bool cmp3(pnt x,pnt y) {return x.a<y.a;}
long long lowbit(long long x) {return (x&(-x));}
void add(long long x,long long val)
{
    for (long long i=x;i<=n;i+=lowbit(i))
        t[i]+=val;
}
long long ask(long long x)
{
    long long ret=0;
    for (long long i=x;i>=1;i-=lowbit(i))
        ret+=t[i];
    return ret;
}
void cdq1(long long left,long long right)
{
    long long mid=left+right>>1;
    sort(p+left,p+mid+1,cmp1);sort(p+mid+1,p+right+1,cmp1);
    long long i=left,j=mid+1;
    while (j<=right)
    {
        while ((i<=mid) && (p[i].b<p[j].b))
        {
            add(p[i].c,1);
            i++;
        }
        p[j].ans+=ask(n)-ask(p[j].c);
        j++;
    }
    for (long long j=left;j<i;j++) add(p[j].c,-1);
}
void cdq2(long long left,long long right)
{
    long long mid=left+right>>1;
    sort(p+left,p+mid+1,cmp2);sort(p+mid+1,p+right+1,cmp2);
    long long i=left,j=mid+1;
    while (j<=right)
    {
        while ((i<=mid) && (p[i].b>p[j].b))
        {
            add(p[i].c,1);
            i++;
        }
        p[j].ans+=ask(p[j].c-1);
        j++;
    }
    for (long long j=left;j<i;j++) add(p[j].c,-1);
}
void cdq(long long left,long long right)
{
    if (left==right) return;
    long long mid=left+right>>1;
    cdq(left,mid);cdq(mid+1,right);
    cdq1(left,right);
    cdq2(left,right); 
}
int main()
{
    scanf("%lld%lld",&n,&m);
    for (long long i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        pos[a[i]]=i;
    }
    for (long long i=1;i<=n;i++)
    {
        f1[pos[i]]=(pos[i]-1)-ask(pos[i]-1);
        ans+=f1[pos[i]];
        f2[pos[i]]=ask(n)-ask(pos[i]);
        add(pos[i],1);
    }
    memset(t,0,sizeof(t));
    for (long long i=1;i<=m;i++)
    {
        scanf("%lld",&x);
        p[i].a=i;p[i].b=pos[x];p[i].c=x;
    }
    cdq(1,m);
    sort(p+1,p+m+1,cmp3);
     for (long long i=1;i<=m;i++)
    {
        printf("%lld\n",ans);
        x=p[i].c;
         ans-=(f1[pos[x]]+f2[pos[x]]);
         ans+=(p[i].ans);
    }
    return 0;
}